Forces of F1=40N at 30 degrees and F2=30N at 150 degrees act on a point. Determine by drawing or by calculation a) F1 + F2 and b) F1 - F2

Please help with answer!!!

Dec 23rd 2008, 08:15 AM

HallsofIvy

Quote:

Originally Posted by kingtim04

Forces of F1=40N at 30 degrees and F2=30N at 150 degrees act on a point. Determine by drawing or by calculation a) F1 + F2 and b) F1 - F2

Please help with answer!!!

Well, have you at least drawn a picture as it suggests? Since the problem says "or by calculation" I would take that to be a sufficient answer.

To do the calculation of F1+ F2 from the picture, you will have a triangle in which one side has length "40" and angle 30 to the x-axis. The second side starts at the end of that, has length "30" and makes an angle 150 degrees with the x-axis and so 150- 30= 120 degrees with the first side. F1+ F2 is the third side of that triangle. You can find its length using the cosine law and the angle using the sine law.

To do the calculation of F1- F2 from the picture, you would have the same first side but F2 goes the opposite direction: 150- 180= -30 degrees so now the angle between the two sides is 30-(-30)= 60 degrees. Again, use the cosine law to find the length of the opposite side and the sine law to find the angle.

You could also do both of these problems by calculating the components of the vectors: F1= (40 cos(30), 40 sin(30)) and F2= (30 cos(150), 30 sin(150)). Add the two vectors by adding the components and subtract by subtracting ythe components.